A Proposal For A Universal Particle Equation
Ian Beardsley
March 10, 2026
ABSTRACT
A Normal force where , which we suggest is invariant, is introduced
that determines a master equation for the proton, neutron, and electron, a kind of Universal
Particle Equation. It is suggested that when we push on a particle we rotate some of its temporal
velocity into spacial velocity and resistance to this rotation is experienced as the normal force
pushing back creating inertia that we experience as mass.
The Proposal
We can form two equations where the proton radius to its mass produces about 1-second [3, 4]:
1.
(Proton Mass)
(Proton Radius)
(Planck Constant)
(Light Speed)
(Gravitational Constant)
1/137 (Fine Structure Constant)
Using equation 6, equations 1 and 2 directly yield [3, 4]:
where . Here we see in equation 3, the cross-sectional area of the proton
is exposed to the normal force, mediated by the 'stiffness of space' as measured by ,
producing the proton mass, .
We suggest equations 1 and 2 are correct because they yield the proton radius closely. They give
it as (by equating them) [3, 4]:
F
n
= h /(ct
2
1
)
t
1
= 1 second
ϕ ⋅
π r
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1 second
2.
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1 second
m
p
: 1.67262E − 27kg
r
p
: 0.833E − 15m
h : 6.62607E − 34J ⋅ s
c : 299,792,458m /s
α :
3. m
p
= κ
p
⋅
π r
2
p
F
n
G
4. F
n
=
h
ct
2
1
5. t
1
= 1 second
κ
p
= 1/(3α
2
)
A
p
= π r
2
p
F
n
G
m
p
The CODATA value from the PRad experiment in 2019 gives
With lower bound , which is almost exactly what we got.
We can see equation 6 may be the case because we get it from Planck Energy ,
Einsteinian energy, , and the Compton wavelength when we
introduce the factor of , which is the golden ratio conjugate, where the golden ratio,
. [3, 4].
I explain this factor by invoking Kristin Tynski, her paper titled: One Equation, ~200 Mysteries:
A Structural Constraint That May Explain (Almost) Everything [2].
Tynski shows that for any system requiring consistency across multiple scales of observation has
the recurrence relation:
Which she says leads to
Whose solution is .
For the proton radius in our computations we will use
"A measurement of the atomic hydrogen Lamb shift and the proton charge radius"
by Bezginov, N., Valdez, T., Horbatsch, M. et al. (York University/Toronto)
Published in Science, Vol. 365, Issue 6457, pp. 1007-1012 (2019).
It has a value of
The 1-second verification follows from the resulting Universal Particle Equation [3]:
Proton: , :
Neutron: :
6. r
p
= ϕ ⋅
h
cm
p
r
p
= (0.618) ⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.8166E − 15m
r
p
= 0.831f m
±
0.014f m
r
p
= 0.817E − 15m
E
p
= hν
p
E
p
= m
p
c
2
λ
p
= h /(m
p
c) = r
p
ϕ
Φ = 1/ϕ = ( 5 + 1)/2 ≈ 1.618
7. scale(n + 2) = scale(n + 1) + scale(n)
8. λ
2
= λ + 1
Φ
r
p
= 0.833f m
±
0.012f m
9. t
1
=
r
i
m
i
⋅
πh
Gc
⋅ κ
i
κ
p
=
1
3α
2
α = 1/137
t
1
=
0.833 × 10
−15
1.67262 × 10
−27
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 6256.33 = 1.00500 seconds
κ
n
=
1
3α
2
t
1
=
0.834 × 10
−15
1.675 × 10
−27
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 6256.33 = 1.00478 seconds
Electron: :
We suggest for the electron may be because it is the fundamental quanta (does not consist
of further more elementary particles).
. (Neutron radius)
. (Classical electron radius)
Equation 9 is a Natural Law. In general we can write the universal law as [4]:
Discussion
It is thought that the proton does not have an exact radius, but that it is a fuzzy cloud of
subatomic particles. As such depending on what is going on can determine its state, or effective
radius. It may be these different sizes are predicted by Fibonacci approximations to [4]. If
such an approximation is given by it could be that the proton radius is as large as
Which it was nearly measured to be before 2010 in two separate experiments. One using
hydrogen spectroscopy, the other electron scattering. In 2010 The recommended CODATA value
was . Then came the shocking 2010 measurement that was 4.2%
smaller using the new Muonic hydrogen result, which was . This resulted in the
famous “proton radius puzzle”.
We might suggest that the proton radius might get still smaller, closer to something using the
Fibonacci approximation of . In which case we would have:
κ
e
= 1
t
1
=
2.81794 × 10
−15
9.10938 × 10
−31
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 1 = 0.99773 seconds
κ
e
= 1
r
n
= 0.84E − 15m
r
e
= 2.81794E − 15m
10. m
i
= κ
i
⋅
π r
2
i
F
n
G
F
n
=
h
ct
2
1
F
n
=
6.62607015 × 10
−34
J·s
(299,792,458 m/s)(1 s)
2
= 2.21022 × 10
−42
N
t
1
= 1 second
ϕ
ϕ ≈ 2/3
r
p
=
2
3
⋅
h
cm
p
r
p
=
2
3
⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.88094E − 15m
r
p
= 0.8775f m
±
0.0051f m
r
p
= 0.84184f m
ϕ ≈ 5/8
In such cases, in equation 1 takes on different ratios between successive Fibonacci ratios.
References
[1]Bezginov, N., Valdez, T., Horbatsch, M. et al. (York University/Toronto)
Published in Science, Vol. 365, Issue 6457, pp. 1007-1012 (2019) "A measurement of the atomic
hydrogen Lamb shift and the proton charge radius"
[2]Tynski, K. (2024). One Equation, ~200 Mysteries: A Structural Constraint That May Explain
(Almost) Everything.
[3]Beardsley, I. (2026) How Physics and Archaeology Point to a Natural Constant of 1-Second,
https://doi.org/10.5281/zenodo.18829259
[4]Beardsley, I. (2026) The Sublime and Mysterious Place of Humans in the Cosmos; A Work in
Exoarchaeolog, https://doi.org/10.5281/zenodo.18715148
r
p
=
5
8
h
cm
p
r
p
=
5
8
(6.62607 × 10
−34
)
(299,792,458)(1.67262 × 10
−27
)
= 0.8258821 × 10
−15
m
ϕ